Question

Find the magnitude of the position vector (in metre) of a projectile after $\sqrt{2}$ seconds of flight whose initial velocity is, $u=30\text{m/s}$ and range is $90\text{m}$. $($Take $g=10{\text{m/s}}^2$$)$

• A. 10
• B. $10\sqrt{7}$
• C. $7\sqrt{13}$
• D. $10\sqrt{13}$

Answer 1

The correct option is D $10\sqrt{13}$

Given, $t=\sqrt{2}s,u=30\text{m/s},R=90\text{m}$
$\Rightarrow R=\frac{u^2\sin 2\theta }{g}$
$\Rightarrow 90=\frac{{30}^2\sin 2\theta }{10}$
$\Rightarrow \sin 2\theta =1$
$\theta =45°$
$u_x=u_y=\frac{30}{\sqrt{2}}\text{m/s}$
Position vector is:
$\overrightarrow{r}=(u_x\times t)\hat{i}+(u_{y}t+\frac{1}{2}{a}_y{t}^2)\hat{j}$
$=(\frac{30}{\sqrt{2}}\times \sqrt{2})\hat{i}+(\frac{30}{\sqrt{2}}\times \sqrt{2}+\frac{1}{2}(-10\left)\right(2\left)\right)\hat{j}$
$=30\hat{i}+20\hat{j}$
$\left|\overrightarrow{r}\right|=\sqrt{{30}^2+{20}^2}=\sqrt{1300}=10\sqrt{13}m$